Table of Contents

Class Notes

Class Work

Homework

Labs

Lab: Constant Speed of a CMV

Lab: Graphing Acceleration

Lab Acceleration Graph Lab 9/14

Lab: A Crash Course in Velocity (Part II) 9/21

Lab: Falling Object 10/5

Projects

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Class Notes

Class Notes: Constant Speed (9/9):
Constant Speed- will always have the same value

• All three of these values always have the same equation (instantaneous, average, and constant)
• Only difference is that for instantaneous, the speed will be really small

4 Types of Motion

• At restConstant Speed
  • Staying still
  • zero velocity
• Increasing speed
  • Speed is not changing
• Decreasing speed

- Acceleration

• Simply means changing speed, so both increasing and decreasing speed are acceleration because they are simply changes in the speed

- Deceleration

• Decreasing speed

- Motion Diagram

• - Show direction of velocity and acceleration only
• - If you are at rest, the velocity is O and acceleration is 0
• - If you walk to the right at constant speed, there could be a motion diagram
• - To show increasing speed, you would have the arrows continuously getting bigger so that it’s obvious that they are changing size; a à moving to the right for positive acceleration
• - To show decreasing speed, you would have the arrows continuously getting smaller; the acceleration would be negative so you would show the a ß to be going to the left
• - Constant speed toward the left would look like this <-- <-- <--
• - These diagrams don’t actually change but the arrows just change direction
• - Quick, easy diagrams that tell about direction

These are pictures of motion diagrams

Ticker Tape Diagram

• At rest, it would be a single dot
• At constant speed, it would be evenly-spaced dots
• Increasing speed will have increasing space with each dot
• Decreasing will have less space in between each one
• These are very precise measurements
• You can’t see direction
• You can’t see where it stopped unless you have more information
• This is generally used when measurements need to be taken

Signs are arbitrary (made up)

• Up and right are positive; down and left are negative
• Sometimes things are moving up a hill and in those cases, you simply tilt the axis so that the positive is going up the hill
• Every vector needs a sign

Kinematics- The study of motion
Class Notes: Graph Shapes (At rest and Constant Speed)

these show x-t graph, v-t graph, and a-t graph at rest and at constant speed

Class Notes: The Big 5 9/12:Kinematics- acceleration (a) = m/s2- rate that velocity is changing- a= Vf-Vi / change in time OR => Vf = Vi + atchange in d= 1/2 (Vi + Vf) tchange in d= vi t + 1/2at squaredVf squared= Vi squared + 2a change d- V= d/t ONLY for average or constant speed--> this does not apply to when there is acceleration- find average speed is literally (V+V) /2 --> ONLY for average speed- 1/2 (Vi + Vf) = d/t
Increasing and Decreasing Speed Graphs

Notes 10/3

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Class Work

Interpreting Graphs (E, F, G)

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Homework

1D Kinematics: Lesson 1 (Method 2a) 9/8

1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
   • Describing Motion with Words
     • - Mechanics- study of the motion of objects
     • - Kinematics- science of describing the motion of objects using words, diagrams, numbers, graphs, and equations; kind of mechanics with goal of making models that our minds can interpret of the way things move
     Scalars and Vectors
     • - Physics is based on math
     • - Useful motion words: going fast, stopped, slowing down, speeding up, and turning
     • - Terms also used: displacement, speed, distance, velocity, acceleration (have strict mathematical meanings)
     • - Scalars- quantities with numerical value only
     • - Vectors- are quantities with numerical value and direction
     Distance and Displacement
     • - Distance- is a scalar quantity that shows how far something traveled in a certain amount of time; doesn’t need direction
     • - Displacement is a vector quantity that explains an object’s change in position in a certain amount of time; needs direction
     Speed and Velocity
     • - Speed- a scalar quantity that shows to how fast an object is moving; something that isn’t moving has 0 speed
     • - Velocity- a vector quantity that shows how fast an object changes position
       • A person stepping back and forth has zero velocity
       • is direction aware
       • Ex. 300 m/h, east
   • Calculating Average Speed and Average Velocity
     • - Average speed= distance traveled/time of travel
     • - Average velocity= change in position/time = displacement/time
     Average Speed vs. Instantaneous Speed
     • - Instantaneous speed- the speed at a given time
     • - Average Speed- average of instantaneous speeds; found distance/time
       • you can literally take the average as well
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
   • We did learn about the difference between speed and velocity in class; speed is the scalar quantity which explains how far something has traveled over the time it took to do so, while velocity is the vector quantity that refers to change in position and is the displacement over the amount of time that it took to do so. The reading helped to clarify these concepts because it worked with example problems, which allowed me to apply these concepts to real-life situations, which was much more helpful than just seeing the definitions and even formulas alone.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
   • I think that I have a pretty good grasp on all of the materials that were discussed in the reading.
4. What (specifically) did you read that was not gone over during class today?
   • We didn't go over the difference between scalar and vector quantities. The reading clarified this though because it explained that the scalar quantities only have numerical value, while vector quantities have numerical value accompanied by a direction.

1D Kinematics: Lesson 2 (Method 2a) 9/9
1. What did you read that you already understood well from our class discussion? Describe at least 2 items fully.
- I read about the ticker tape diagrams, which are basically just strips of paper that have dots running across them. The key to ticker tape diagrams though is that you need to know and understand how to read them and understand what they are indicating. When the dots are evenly spaced apart, that means that there is a constant speed of the object moving, pulling the paper through the machine. On the other hand, if the distance between the dots is increasing, that means that the speed is increasing and if the space between them is decreasing, so is the speed. This shows acceleration, which is a change in the speed either up or down and in the velocity or change in position. The only problem with these ticker diagrams is that they don't show the direction that the object is going in, which Vector diagrams do show. Vector diagrams are basically diagrams that show both the acceleration and the velocity of the object through a series of arrows. When the arrows are longer or shorter, this shows that there is a change in speed and/or velocity, similar to the differing spaces in-between the dots. The vector diagrams do show the direction though, with arrows going different ways. For example, an object going east would have an arrow going -->. The arrow representing velocity always goes in the direction of the object, while the arrow representing acceleration may be going the opposite way if the object is slowing down. This is because there is a negative acceleration when an object is slowing down.
2. What did you read that you were a little confused/unclear/shaky about from class, but the reading helped clarify? Described the misconception you were having as well as your new understanding.
- I found the concepts that we learned in class to be pretty clear. Though, these readings did definitely help in reviewing the material and giving some good examples, as well as diagrams. The animation diagram really helps to understand the vector diagrams because it shows that as the object's acceleration changes, the arrow moves to face the opposite direction.
3. What did you read that you still don't understand? Please word these in the form of a question.
- I think everything seems pretty clear between my class notes and the reading.
4. What did you read that was not gone over during class today?
- We didn't completely learn about what we will be using these diagrams for, besides showing speed and velocity. The reading indicates that forces can be shown through these diagrams, and they are very important for showing sizes of a quantity. The reading also gave examples of ticker tape diagrams and turned them into problems to solve, which was helpful in understanding that when the dots are close, it is going slower, while when they are farther apart, the object is speeding up.

1D Kinematics: Lesson 3 (Method 2a) 9/13
1. What did you read that you already understood well from out class discussion? Describe 2 items fully.
- Basically, it is important to realize that acceleration is not what many people think it is, going fast, but it is rather about the change in the velocity. While some people think that something has acceleration if it is going fast at a constant speed, in reality, it has no acceleration because there is no change in speed. It is also important to note that when the acceleration is slowing down, it is actually going in the opposite direction of the motion, and in the case that it is going opposite the motion, the acceleration becomes negative, rather than positive, which would be increasing speed.
2. What did you read that you were a little confused about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
- It was helpful to see the formula again and to revisit it. I know that the ratio also would have to be velocity units per time units, as the reading explained. The examples at the end showed how to apply the changes in the velocity over the time, which helped me when working on the problems in the textbook.

3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
- I think that I understand everything in the reading.
4. What (specifically) did you read that was not gone over during class today?
- I didn't know that acceleration was a vector quantity, meaning that it required a direction besides a number quantity. Direction is extremely significant in understanding acceleration because whether it is going north, east, south, or west, understanding the direction will help define if the acceleration is positive or negative. We also didn't learn about free-falling objects very much and how in every second that they are accelerating at a constant rate, the speed is increasing, and therefore, it is traveling a greater length.

1D Kinematics: Lesson 4 (Method 2a) 9/15
Lesson 3
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
- I already knew from our class discussion that a position time graph showing a constant, positive velocity would be a single linear line. I also knew that a position time graph with a changing velocity, or acceleration, would create a curved line. We learned about this in our lab...acceleration will always create this kind of curved shape line because the slope is constantly changing, due to the constant change in displacement and position from the changing velocity.
- I also definitely knew about how to find the slope of the graph. This simple formula is very important to understanding all the parts of the graph. We know that slope= change in y/change in x and can be found by choosing any two points on the line of either the specific line being chosen or in the case of a curved line, picking two points on the line tangent and plugging them into the formula of (y2-y1)/(x2-x1).
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
- The animation helps to explain visually the way that the position time graph works. This also explains and clarifies that the position time graph is referring to the specific position, rather than the speed. It explains that if this were a velocity vs. time graph, we would not see the lines intersecting when the cars were next to each other because the lines would realistically be two horizontal lines, being they have constant velocities. But we do see on this animation that the position time graph shows the different speeds through steepness, while also depicting their position.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
- I feel like I understand everything.
4. What (specifically) did you read that was not gone over during class today?
- I think that we went over pretty much everything on the lesson.
Lesson 4
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
- I knew about how in a v-t graph, when there is a constant velocity, the graph would show a straight horizontal line parallel to the x-axis. When a car has an acceleration on the v-t graph though, the line becomes a straight increasing line. This is because the velocity is increasing as the time goes on and position is changing. It is also important to note that if the line crosses over the x-axis, that means that the object has changed directions.

- The shape of the v-t graph is really important to finding the acceleration, or speed change, in the object. If there is no change in velocity, then the slope of the velocity time graph line would be zero, which would also be an indication of the acceleration rate, which would be zero as well because velocity in that case would be a constant speed.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
- It is helpful to see these graphs below because it explains visually how to interpret the graphs. When the speed is increasing, the line is positively going up, while when it is slowing down, the line goes negatively down. We see it under the x-axis when the object is moving in the opposite direction, because as we learned earlier, velocity is a vector quantity, meaning that direction is important and counts. When the object is approaching the x-axis, it is slowing.

3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
-I feel like I understand everything.
4. What (specifically) did you read that was not gone over during class today?
- Although we did briefly mention finding the area in class on a v-t graph, we didn't go into enough detail to really understand the concept. This graph can be used to show displacement of an object. Almost any shape can be taken (rectangle, triangle, trapezoid) and the way to find displacement is simply to find the area of the shape. I was definitely already aware of the area formulas for rectangles, triangles, and trapezoids, although for trapezoids, another formula can be used. For the trapezoid, it is also possible to find the area of the triangle and rectangle separately within it, which is really just applying methods that we already know.

Method 1: Lesson 5 10/3
Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance. All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s known as the acceleration of gravity (sole influence of gravity) and this value has the g symbol to denote it. A curved line on a position versus time graph signifies an accelerated motion and since a free-falling object is undergoing acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. Observe that the line on the graph is a straight, diagonal line. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is vf= g * t where g is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. Ex. At t 6 s
vf (9.8 m/s2) * (6 s) 58.8 m/s. The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by the formula. d = 0.5 * g * t2 where g is the acceleration of gravity (9.8 m/s/s on Earth. Ex. At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus, the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass.

Method 1: Lesson 1 A and B

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Labs

Lab: Constant Speed of a CMV

Partner: Dani Rubenstein

Objective: What is the speed of a Constant Motion Vehicle (CMV)?
Available Materials: Constant Motion Vehicle, Tape measure and/or metersticks, spark timer and spark tape
Hypothesis: The speed of the car is approximately 40 cm/s. I approximated this based on the idea of how many centimeters I cover in a minute which would be probably about 60 or more. I figured the car would move slower. The measurements can be taken with a ruler and approximated to the hundredth of a centimeter. Position-Time graphs most likely show where the object stands at a specific time.
Data:
Position-Time Table Data for CMV:

this graph shows the position of the car at different time intervals

Analysis of Data:
The R-squared value shows how accurate the data fits the trend-line, which appears to be approximately .99, which looks about correct considering it needs to be as close to 1, or 100% as possible in order to be accurate. This graph shows the position of the car at particular time intervals which were measured on the piece of ticker tape which went through the spark timer. Because this is a position time graph, our graph shows that the position was constantly changing as time passed and speed was increasing.


Discussion questions

1. Why is the slope of the position-time graph equivalent to average velocity?
   • The slope of the position-time graph is equivalent to the average velocity. This is because the slope is equal to the change in y (position) over the change in x (time) and so is the velocity, being that it is equal to the total change in the distance over the total change in time.
2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?
   • Average velocity is the average of all of the different recorded speeds, while instantaneous velocity is the speed at a given time. We are assuming that there aren't any extreme velocity values, which could change the overall average. We're also assuming by finding the average velocity that the car is moving at a constant speed.
3. Why was it okay to set the y-intercept equal to zero?
   • At the time of zero, the car was not moving and was equal to the position of zero. Therefore, both time (x-coordinate) and position (y-coordinate) should
     theoretically be at zero, making the trend line cross the origin.
4. What is the meaning of the R2 value?
   • The R2 value is used to show how accurately the trend line fits the data points, in a percentage. We use it to correctly evaluate the data. The closer it is too 1, the more accurate the trend line.
5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
   • I would expect it to lie below my graph because the time variable is the same, although the position variable is lower. The two graphs do start at the origin because the time starts at zero and so does the position, although the speeds are different, making the slower speed graph to appear lower and less steep. This is because it has a smaller slope value, due to the lower speed.

Conclusion
Our resulting line was y=61.663x. The hypotheses that I expected were that: the CMV would move about 40 cm/s, that millimeters were the most precise form of measurement, and that a position time graph explains the position of an object and its direction at a certain time. The CMV ended up moving at about 61.663 cm/s. Measurements can be measured more precisely than millimeters, and can be measured to the hundredth of a centimeter. A position-time graph explains the speed of an object, which is shown through the relationship of time and position.There could have been many sources of error. These include possibilities of the ruler shifting, the limiting measurement tools, which only allowed us to measure to the hundredth’s place, and the distorted point of view caused by the ruler’s thickness. In the case of redoing this lab, I would try to minimize these issues by requesting measuring tape to hold down the ruler and keeping it from shifting. I would also use a more precise piece of measuring equipment that would allow me to measure further than to the hundredth’s place. I would use measuring tape so that the point of view issue wouldn’t be a problem, considering that the tape is flat, rather than thick and high (the reasons that could have resulted in inaccuracies).

Lab: Graphing Acceleration

At Rest

Constant Speed

Change in Direction

Constant Speed Fast

Position-Time Graphs explain the position vs. the time. The slope of them is equal to the velocity, and therefore, shows how fast you are going. When the line on this graph is steeper, it shows a faster moving object. Acceleration objects show a curved line on x-t graphs.
Velocity-Time Graphs show the velocity at specific time intervals. When velocity is constant, it will stay a horizontal line. When there is an acceleration in the velocity, the v-t graph will show a positive or negative line. Acceleration-Time Graphs show if the velocity is changing or not. These graphs will always be horizontal and only show change when there is change in the speed.


1. How can you tell that there is no motion on a…
position vs. time graph--> it is at the starting point
velocity vs. time graph--> horizontal line at zero
acceleration vs. time graph --> horizontal line at zero

2. How can you tell that your motion is steady on a…
position vs. time graph--> constant slope
velocity vs. time graph--> horizontal line
acceleration vs. time graph--> horizontal line

3. How can you tell that your motion is fast vs. slow on a…
position vs. time graph--> steeper slope indicates a faster moving object
velocity vs. time graph--> y-value of the line would be greater of the faster moving object than the slower one
acceleration vs. time graph--> the speed stays the same so there is no acceleration, keeping it at zero

4. How can you tell that you changed direction on a…
position vs. time graph--> the slope will become negative if you turned around, or positive if it was previously negative
velocity vs. time graph--> reaches zero and still decreases under x-axis
acceleration vs. time graph--> this graph doesn't show direction, only the change in speed

5. What are the advantages of representing motion using a…
position vs. time graph--> shows the speed of an object in a particular position
velocity vs. time graph--> slope shows the increase or decrease in velocity
acceleration vs. time graph--> slope shows the change in speed, if there is any

6. What are the disadvantages of representing motion using a…
position vs. time graph--> acceleration isn't shown, because the changes in speed aren't shown
velocity vs. time graph--> this fails to show position
acceleration vs. time graph--> this doesn't show the actual velocity, only the changes within the speed

7. Define the following:
No motion--> an object is at rest and so, velocity, position, and speed all remain at zero
Constant speed--> there is a continuous velocity and acceleration remains at zero because there is no fluctuation in the speed

Lab Acceleration Graph Lab 9/14

Partner: Dani Rubenstein
Objectives:

• What does a position-time graph for increasing speeds look like?
• What information can be found from the graph?

Available Materials:
- Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
Hypothesis:
- Due to the increasing acceleration, or change in speed, the graph would gradually get steeper.
Procedure:

This video shows the procedure that we used in this lab. We attached ticker tape to the end of the cart and let it go down the ramp, which was created by being placed on the edge of a textbook. The ticker tape went through the Spark Timer, which left a series of dots along the ticker tape, with distance between the dots indicating distance traveled per tenth of a second.
Data and Graphs:
Data Table of Decline chart:

This graph shows the position of the cart at different time intervals

Analysis:
a) Interpret the equation of the line (slope, y-intercept) and the R2 value.
- Equation of the Line: y= 13.254x^2 + 8.1048x
Because we learned in class that the equation of the line y=Ax^2 + Bx is the same as the formula of change in distance = (1/2a) (t^2) + (Vi)(t), this implies that 13.254 x 2 is the acceleration, which equals 26.508 and that 8.1048 is equal to the initial velocity of the object, as shown in the graph.
- On our graph, our R2 value was 0.99, or 99%. The trend-line that we used was polynomial though, rather than linear, because when we used a linear trend-line, the R2 value was approximately 96%, indicating that the linear trend-line did not fit the data. The polynomial fit the data better because our data is shaped in a curve-like form.
b) Find the instantaneous speed at halfway point and at the end.
- The instantaneous speed at the halfway point was 36.67 cm/s. The instantaneous speed at the end was 93.33 cm/s.

this shows the x-t graph with lines going through both the half point and the endpoint to find instantaneous speeds halfway and at the end

c) Find the average speed for the entire trip.

Discussion Questions:
1. What would your graph look like if the incline had been steeper?
- The curve would have been steeper and would have gone up higher sooner. The y-values would have been greater for each x-value, so the curve would have appeared much steeper in less time.

Graph if Incline was Steeper

2. What would your graph look like if the cart had been decreasing up the incline?
- It would be a curve starting out steeper and it gets flatter as time increases because the acceleration slows as the car goes up the ramp.

3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
- The instantaneous speed at the halfway point was only 36.67 cm/s, while the average speed of the entire trip was 30.35 cm/s. Because the instantaneous speed was taken from the midpoint of the graph, it is similar to the average speed of the entire trip. The average speed is the total distance over the total time, while the midpoint was instantaneous, although it was in the middle of the acceleration's increase, and therefore they share similar scalar quantities.
4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
- The slope of the tangent line is equal to the instantaneous speed because when a line is drawn tangent to that point, this shows the change in distance over the change in time. The line is a curve, so we must use the values of each particular point to find the speed in only that instant, which would be impossible to do otherwise.
5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible

This graph shows the velocities of the cart, taken from the graphs above, finding velocity from using change in distance over change in time

Discussion:
Although I had originally thought that the line would simply be a linear positively increasing line, this hypothesis was not correct. Because the increase in acceleration is gradual, the shape of the graph actually forms a curve, and therefore takes the form of a polynomial. We figured this out through the R2 value, which proved that the data fit the polynomial line better, having a 99% fit. As the time passed, the cart was covering greater amounts of distance, which makes sense, because its speed was accelerating positively. We found that the
There were multiple places where error could have affected our results. My partner and I did not start from the first dot on the ticker tape, which could have contributed to the error. The data taken down could have been slightly off, for we measured the ticker tape by hand, which left a lot of room for error because we can only estimate so far and so accurately. Doing this lab in the future, I would use different systems of measurement for taking down either the distance between the ticker tape dots, or use a different method for finding the velocity of the cart going down the ramp that is more accurate.

Lab: A Crash Course in Velocity (Part II) 9/21

Partners: Dani, Ali, Caroline
Objectives: Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.

PART A
Find another group with a different CMV speed. Find the position where both CMV’s will meet if they start at least 600 cm apart, move towards each other, and start simultaneously.

At this point in the lab, we tried to find the time that would pass before the two CMVs collided. We found that the time that would pass when they collided was 6.038 seconds. At this time, the yellow car will have moved 229.215 cm, while the blue car will have moved -372.321 cm. The distance for the blue car is negative because this is showing direction, and because both the cars are going in opposite directions, one must be negative, although the signs are arbitrary.

PART B
Find the position where the faster CMV will catch up with the slower CMV if they start at least 1 m apart, move in the same direction, and start simultaneously.

At this point in the lab, we tried to find the time and position at which the blue car would catch up to the yellow car. We found that the time would be 4.219 seconds, at which the blue car will have traveled 260.156 cm.

Available Materials: Constant Motion Vehicle, Tape measure and/or metersticks, Masking tape (about 30 cm/group), Stop watch, spark timer and spark tape
Procedure:
Part A:

This video shows the two cars colliding at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
Part B:

This video shows the faster of the two cars catching up to the slower of the two cars at a certain point. We then proceeded to mark the place where they met, proving our calculations to be approximately correct.
Data:
Results from Problem 1:

Results from Problem 2:

Analysis:
PART A
Percent Error:

Percent Error for Blue Car Results

Percent Error for the Yellow Car Values

Percent Difference:


Analysis Part A: Our percent error ranged from 0.26% to 0.91% for the blue car. For the yellow car, our percent error ranged from 1.15% to 2.19%, which shows that the yellow car values for Part A were slightly more off, although these values are all relatively low, indicating that the error was very slight. Percent differences were also very slight, as would be expected because the error was very slight. The values for the blue car ranged from 0.002% to 0.33%, while for the yellow car they ranged from 0.004% to 0.56%.

PART B
Percent Error:


Percent Difference:

Analysis Part B: Our percent error ranged from 0.23% to 11.64%, showing especially high error in the first couple of trials. We probably were doing something slightly different which we might have fixed in our later trials because the error was decreased to 0.23%. As expected, our results for percent difference ranged from 0.20% to 11.61%, showing again that our error was most likely present in the earlier trials and then got slighter in the later trials.

Discussion questions

1. Where would the cars meet if their speeds were exactly equal? Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time.
   If the car speeds were exactly equal and going towards each other, they should theoretically meet exactly at the halfway point, which would be at 300 cm. If the car speeds were exactly equal and going away from each other, they should never meet because they are going in opposite directions.
2. 
   

   Crashing Position Time Graph

3. catching up Position Time Graph

4. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?

this shows the velocities when the cars collide

this graph shows the velocities of the cars when the faster meets up with the slower

There is no way to find the points when they are at the same time place at the same time because this graph doesn't show position and the cars actually started out in different positions as well, which this graph fails to represent.

Conclusion:
For the most part overall, the predictions from our equations approximately matched our data from our trials. We said that the two cars would meet when the yellow car had traveled 229.215 cm and the blue car had traveled 372.321 cm for the first problem. Our results did range slightly, although all of the data was very close to these numbers. In Part A, for the blue car, the percent difference ranged from 0.002% to 0.33%, while in the yellow car the percent difference ranged from 0.004% to 0.56%. These calculations did allow us to clearly understand that the blue car was obviously traveling faster than the yellow car, considering in the same amount of time, it had traveled more distance, giving it a higher speed. For the second problem, we calculated that the blue car would catch up to the yellow one once the yellow car had reached 260.156 cm. All of the data results that we got were considerably close to this number, with a few outliers. In Part B, the percent difference ranged from 0.20% to 11.61%. Some error was definitely present within the lab. This could have been from us not being able to specifically let the cars go on the track at the exact time, or turning them on at slightly different time intervals. We did put our cars on metal tracks to keep them going straight, although they could have been veering slightly which could have affected friction against the metal rims, and therefore their speeds. When we measured the distances, we only rounded to the thousandth's place, so they could have been more accurate. In our first calculation, the sum of our two distances didn't equal exactly 600, which could have also contributed to the error. Using more accurate measurement tools and more advanced technology could help to improve the results we got in the future. If we had some sort of equipment that could find the exact place at which the two cars were colliding at, this would be helpful; this would also be helpful when the one car is catching up. If we had some kind of motion detector that could sense where the two cars either collided or were next to each other, we could probably incorporate the computer to get more accurate results.

Lab: Falling Object 10/5

Partner: Dani Rubenstein
Objective: What is the acceleration of a falling body?
Materials: Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meterstick.
Procedure: We taped ticker tape to a small 100 g circle weight after putting the ticker tape through a spark timer. We then held the spark timer and the weight over the railing in the main lobby. From there, we dropped the weight, which sent the ticker tape through the spark timer, until the weight hit the ground. We then measured the distances from the beginning of the ticker tape's dots to each dot. We recorded these distances and used excel to graph our results.
Hypothesis: The approximate value of the acceleration of a free-falling object should be -9.8 m/s/s. The graph will look like a velocity-time graph in which it is accelerating towards the origin. The position-time graph should have a curve upwardly to the right because the object has an acceleration, although it is decreasing.
Data

The table to the left shows the Position-Time Data and the table to the right shows the Instantaneous Speed-Mid-time Results from our data

Percent Error

This is our velocity-time graph of the free-falling object

Analysis: This is our velocity-time graph. We set a trend-line for our data, which was best fit by the linear shape. The r-squared value turned out to be 0.9966, which is considerably close to 1. Therefore, our r-squared value is good and shows that we have good data, which is somewhat accurate. This graph shows that the increase in velocity was constant, because the data points fit into a near-linear line. The slope is 710.65, which is equal to the experimental acceleration. We found this through the equation of our trend-line, y=710.65x+24.609. This is slightly off, considering the value for the constant acceleration should be 981 cm/s/s theoretically.

This it the Position-Time Graph, which shows the data of our free-falling object

Analysis: This is our position-time graph. We set a trend-line for our data, which was best fit by the polynomial shape. The r-squared value turned out to be 0.9999, which is extremely close to 1. Our r-squared value shows that our results were very good, for they fit the line well. We can see here that the position was constantly changing and as more time went by, the weight traveled at a faster velocity. From this equation, we know that double of the value found before the x-squared is equal to the acceleration of the object, through substitution of the equation d=1/2at2 + vit. Therefore, the value is 711.34, which is very close to the value that we got for the acceleration in our velocity-time graph, 710.65.

Discussion Questions
1. Does the shape of your v-t graph agree with the expected graph? Why or why not?
No, the shape of my v-t graph does not agree with the expected graph. This is because I had previously expected that the velocity-time graph would show a negative slope, while the one that we have from our data actually shows a positive slope.
2. Does the shape of your x-t graph agree with the expected graph? Why or why not?
The position-time graph agrees with what I expected because I had previously expected that as more time passes, the position will increasingly change, considering that the the weight was constantly changing position from its release from at rest then freefalling to the floor.
3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
Our results came to 710.65 cm/s/s. The average of the class was actually 839.417. We found that our percent difference was 15.34%. Although this value does still fall within the 20% difference range, the significant difference does signal that our results were slightly off.

Percent Difference Calculations

4. Did the object accelerate uniformly? How do you know?
Looking at the velocity-time graph, we do see that the line is almost linear and shows a clear continual increase. We know that the acceleration is approximately uniform because the slope of the velocity-time graph is equal to the acceleration, and because there is an almost-constant slope represented in the trend-line, we can assume that the rate was approximately uniform. We know that the line was very fitting to the data because the r-squared value was 0.9966, which is very close to 1, further supporting the theory that this line correctly represents the constant acceleration.
5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
Latitude causes gravity to be higher than it should be; there is a greater force of gravity at the poles of the Earth than at the equator. Altitude would cause gravity to be lower than it should be.

Conclusion
Originally, in creating a hypothesis for this lab, we had concluded that the data would probably form a linear line that would be increasing velocity towards the origin in the velocity time graph. This hypothesis ended up being incorrect, because the velocity-time graph actually provided data showing that the velocity was increasing away from the origin. I hypothesized that the position-time graph would look like it was increasing away from the origin in an upward curve, which it did end up looking like because as time passed, the position was changing increasingly more, for the acceleration was constant. The slope of our line was 710.65 cm/s/s, which is equal to the acceleration. Because our trend-line was linear, this implies that throughout our data, the slope was approximately constant, implying that the acceleration was also approximately constant. Though, the slope was theoretically supposed to be approximately 981 cm/s/s, so it was definitely off by about 15.34%, although because this is still in the 20% difference range, the results are somewhat relative. Our percent error was significantly large, being 27.56%, so that shows that there was definitely some places in which our results were off. There was some error that probably occurred throughout the lab and the biggest problem came from the friction caused by the spark timer. When releasing the tape through the timer, each time it made a dot, there was friction between the paper and the ink placed on it, which could have slowed the tape moving through. Air resistance could have played a very minimal role in messing up the results, as could have the jolt of the hand releasing the weight from rest. When we were measuring the distance of each dot on the ticker tape, the measuring tape or the spark tape could have moved slightly, throwing off our data. In order to change this experiment slightly to address the errors, it would definitely be helpful to have a more precise way of measuring the change in position like an electronic program that could pick up how much of the tape had passed without actually making contact with it, and recording the results on a computer. This would also solve the problem of exactness when measuring, for we wouldn't have to measure the spark tape using a measuring tape.

Projects

Egg Drop Project

Results:
The final mass of the device was 73.72. The egg mass was 58.51, so our total mass including both the egg and the device was 132.23.

Analysis:
Our egg did have a slight crack in it and was bleeding. I believe that this was because there wasn't enough protection around the actual egg itself to protect it from cracking. The basket that we made of straws was helpful because it obviously protected the egg and because the egg was suspended in the middle of it within another contraption, it provided space for the egg to slightly move up and down to break the fall. There was paper on top of the egg and underneath it to provide cushioning, which also helped to break the fall. The parachute on top helped as well because it used air pressure to slow down the velocity while going down.
What would you do differently?
In the case that we made another model, we would probably stick with the idea of making a cone out of paper with a parachute on top. This is because the cone would have given the egg more space to break the fall. I also would have covered the egg in tin foil to give it more protection, almost as another shell, which I believe would have prevented it from cracking, at least as much as it did.